Question: There are four complex numbers $z$ such that
\[z \overline{z}^3 + \overline{z} z^3 = 350,\]and both the real and imaginary parts of $z$ are integers.  These four complex numbers are plotted in the complex plane.  Find the area of the quadrilateral formed by the four complex numbers as vertices.
Solution: Let $z = x + yi,$ where $x$ and $y$ are integers.  Then
\begin{align*}
z \overline{z}^3 + \overline{z} z^3 &= z \overline{z} (z^2 + \overline{z}^2) \\
&= |z|^2 ((x + yi)^2 + (x - yi)^2) \\
&= (x^2 + y^2)(x^2 + 2xyi - y^2 + x^2 - 2xyi - y^2) \\
&= (x^2 + y^2)(2x^2 - 2y^2) = 350,
\end{align*}so $(x^2 + y^2)(x^2 - y^2) = 175.$

Since $x^2 + y^2$ is positive, $x^2 - y^2$ is also positive.  So we seek the ways to write 175 as the product of two positive integers.  Also, $x^2 + y^2 > x^2 - y^2,$ which gives us the following ways:
\[
\begin{array}{c|c|c|c} 
x^2 + y^2 & x^2 - y^2 & x^2 & y^2 \\ \hline
175 & 1 & 88 & 87 \\
35 & 5 & 20 & 15 \\
25 & 7 & 16 & 9
\end{array}
\]The only possibility is $x^2 = 16$ and $y^2 = 9.$  Then $x = \pm 4$ and $y = \pm 3,$ so the four complex numbers $z$ are $4 + 3i,$ $4 - 3i,$ $-4 + 3i,$ and $-4 - 3i.$  When we plot these in the complex plane, we get a rectangle whose dimensions are 6 and 8.

[asy]
unitsize(0.5 cm);

pair A, B, C, D;

A = (4,3);
B = (4,-3);
C = (-4,-3);
D = (-4,3);

draw(A--B--C--D--cycle);

dot("$4 + 3i$", A, NE);
dot("$4 - 3i$", B, SE);
dot("$-4 - 3i$", C, SW);
dot("$-4 + 3i$", D, NW);
[/asy]

The area of this rectangle is $6 \cdot 8 = \boxed{48}.$